Question

The pH of a solution prepared by mixing 69.5 mL of 0.125 M Mg(OH)2 and 197.6 mL of 0.125 M HCl is ________. Submit your answer in the following format: 2 decimal places.

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Answer 1

milimole (mmoll : concentration (M) X volume(mil initial mmol of M9COH)2 = 0.125 M X 69.5 m2 = 8.6875 mmol initial mmol of HCl = 0.125 M X 197.6 mL = 24.7 mmol After mixing both final volume of solution = 69.5 ml + 197.6 mL = 267.1ml Neutralization reaction Mg(OH), caql + 2HCl(aq) → MgCl2 lag) + 2 H₂0 (1) I mmol of Mg(OH), neutralize e mmol of HCL 8.6875 mmol of Mg 10H2 neutralize 8 x 8.6875 mmol of Hel Used mmol of HCl = 2x 8.6 875 = 17,315 mmol Remaining mmol of HCL : initial mmol - used mmol of HQ of HCl = 24.7 mmol - 17.375 mmol = 7.325 Mmo! concentration of remaining HCl : Remaining mmol final volume = 7.325 mmol = 0.0274 M 267.1ml Hol is strong acid, so it present in its ion form cy it depsociates completely in aqueoue solution Hel cag) H+ (aq) + cr-cagl After dissociation 0.0274 M 0.0274M [H+] = 0.0274 M pH = -log (H+] - - 109 (0.0274) = 1.56 pH = 1.56 Answer Piecue Rate